Forward Kinematics:

(1)

1. Write out the entries of a rotation matrix $^A_BR$.

Solution:
\hat{X}_B\cdot\hat{X}_A  \hat{Y}_B\cdot\hat{X}_A   \hat{Z}_B\cdot\hat{X}_A
\hat{X}_B\cdot\hat{Y}_A  \hat{Y}_B\cdot\hat{Y}_A   \hat{Z}_B\cdot\hat{Y}_A
\hat{X}_B\cdot\hat{Z}_A  \hat{Y}_B\cdot\hat{Z}_A   \hat{Z}_B\cdot\hat{Z}_A


2. Express \hat{X}_B=[0, 1, 0]^T in frame \{A\}.

Solution:
\hat{Y}_B\cdot\hat{X}_A
\hat{Y}_B\cdot\hat{Y}_A
\hat{Y}_B\cdot\hat{Z}_A


3. Now calculate ^B_AR:

Solution:
\hat{X}_B\cdot\hat{X}_A  \hat{X}_B\cdot\hat{Y}_A   \hat{X}_B\cdot\hat{Z}_A
\hat{Y}_B\cdot\hat{X}_A  \hat{Y}_B\cdot\hat{Y}_A   \hat{Y}_B\cdot\hat{Z}_A
\hat{Z}_B\cdot\hat{X}_A  \hat{Z}_B\cdot\hat{Y}_A   \hat{Z}_B\cdot\hat{Z}_A

4. Verify that the solution from (2) indeed reduces to [0, 1, 0]^T when expressed in coordinate system B. Tip: Express multiplications as scalar products and remember that the scalar product of orthogonal vectors is 0 and that of parallel vectors is 1.

Solution:

The solution consists of a 1x3 vector that is obtained by multiplying the 1x3 vector from (2) with each row of ^B_AR.  The first entry is:

[\hat{X}_B\cdot\hat{X}_A   [\hat{Y}_B\cdot\hat{X}_A+
\hat{X}_B\cdot\hat{Y}_A  .  \hat{Y}_B\cdot\hat{Y}_A+
\hat{X}_B\cdot\hat{Z}_A]    \hat{Y}_B\cdot\hat{Z}_A]

The first vector is ^A\hat{X}_B and the second vector is ^A\hat{Y}_B. As they are orthogonal, their scalar product is zero.

The second entry is:

[\hat{Y}_B\cdot\hat{X}_A  [\hat{Y}_B\cdot\hat{X}_A
 \hat{Y}_B\cdot\hat{Y}_A . \hat{Y}_B\cdot\hat{Y}_A
 \hat{Y}_B\cdot\hat{Z}_A]  \hat{Y}_B\cdot\hat{Z}_A]

Both vectors are identical and their scalar product is one.

The third entry is:

[\hat{Z}_B\cdot\hat{X}_A   [\hat{Y}_B\cdot\hat{X}_A
 \hat{Z}_B\cdot\hat{Y}_A .  \hat{Y}_B\cdot\hat{Y}_A
 \hat{Z}_B\cdot\hat{Z}_A]   \hat{Y}_B\cdot\hat{Z}_A]

The first vector is ^A\hat{Z}_B and the second vector is ^A\hat{Y}_B. As they are orthogonal, the scalar product is zero.

(2)

Consider two coordinate systems \{B\} and \{C\}, whose orientation is given by the rotation matrix $^C_BR$ and have distance $^CP$. Provide the homogenous transform ^C_BT and its inverse ^B_CT.

Solution:

[
    ^C_BR    ^CP
	0   0   0   1]


